Let $f(x) = 8x^{2}+8x-1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Explanation: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $8x^{2}+8x-1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 8, b = 8, c = -1$ $ x = \dfrac{-8 \pm \sqrt{8^{2} - 4 \cdot 8 \cdot -1}}{2 \cdot 8}$ $ x = \dfrac{-8 \pm \sqrt{96}}{16}$ $ x = \dfrac{-8 \pm 4\sqrt{6}}{16}$ $x =\dfrac{-2 \pm \sqrt{6}}{4}$